Understanding the Problem

The problem provides two trigonometric equations involving positive acute angles A and B:

1. \(\cos 2B = 3 \sin 2A\)

2. \(3 \sin 2A = 2 \sin 2B\)

We are asked to find the value of the expression \((A + 2B)\). Since A and B are positive acute angles, we know that \(0 < A < \pi/2\) and \(0 < B < \pi/2\). This implies \(0 < 2A < \pi\) and \(0 < 2B < \pi\).

Analyzing the Given Equations

From the two given equations, we can equate the expressions that are equal to \(3 \sin 2A\):

\(\cos 2B = 2 \sin 2B\)

Assuming \(\cos 2B \neq 0\), we can divide both sides by \(\cos 2B\):

\(\frac{\cos 2B}{\cos 2B} = \frac{2 \sin 2B}{\cos 2B}\)

\(1 = 2 \tan 2B\)

\(\tan 2B = \frac{1}{2}\)

Since B is a positive acute angle (\(0 < B < \pi/2\)), the angle \(2B\) must be in the range \(0 < 2B < \pi\). As \(\tan 2B = 1/2\) is positive, \(2B\) must be in the first quadrant, i.e., \(0 < 2B < \pi/2\). This is consistent with B being acute.

Connecting to the Expression (A+2B)

We need to find the value of \(A+2B\). Let's consider the trigonometric identity for the tangent of the sum of two angles, \(\tan(X+Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y}\). In our case, \(X=A\) and \(Y=2B\).

Consider the special case where the sum of two positive acute angles is \(\pi/2\). If \(X+Y = \pi/2\) where \(0 < X < \pi/2\) and \(0 < Y < \pi/2\), then \(X = \pi/2 - Y\). Taking the tangent of both sides:

\(\tan X = \tan(\pi/2 - Y)\)

\(\tan X = \cot Y\)

\(\tan X = \frac{1}{\tan Y}\)

So, \(\tan X \tan Y = 1\). This property holds for two positive acute angles summing to \(\pi/2\). If one of the angles is \(2B\) (which we know is in \((0, \pi/2)\)) and the other is \(A\) (which is acute, \(0 < A < \pi/2\)), and their sum \(A+2B = \pi/2\), then the property \(\tan A \tan 2B = 1\) must hold.

We have found \(\tan 2B = 1/2\). If \(A+2B = \pi/2\), then we must have:

\(\tan A \times \frac{1}{2} = 1\)

\(\tan A = 2\)

Verifying Consistency with Original Equations

Now, we need to check if the given equations are consistent with \(\tan A = 2\) and \(\tan 2B = 1/2\), assuming A and B are acute.

If \(\tan A = 2\) and A is acute, we can find \(\sin 2A\) using the identity \(\sin 2A = \frac{2 \tan A}{1+\tan^2 A}\):

\(\sin 2A = \frac{2(2)}{1+(2)^2} = \frac{4}{1+4} = \frac{4}{5}\)

If \(\tan 2B = 1/2\) and \(2B\) is in \((0, \pi/2)\), we can find \(\sin 2B\) and \(\cos 2B\). Consider a right triangle with opposite side 1 and adjacent side 2, the hypotenuse is \(\sqrt{1^2+2^2} = \sqrt{5}\).

\(\sin 2B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{5}}\)

\(\cos 2B = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}}\)

Let's substitute these values into the original equations:

Equation 1: \(\cos 2B = 3 \sin 2A\)

Substituting the values: \(\frac{2}{\sqrt{5}} = 3 \left(\frac{4}{5}\right) = \frac{12}{5}\). This simplifies to \(2 \times 5 = 12 \sqrt{5}\), which is \(10 = 12\sqrt{5}\).

Equation 2: \(3 \sin 2A = 2 \sin 2B\)

Substituting the values: \(3 \left(\frac{4}{5}\right) = 2 \left(\frac{1}{\sqrt{5}}\right)\). This simplifies to \(\frac{12}{5} = \frac{2}{\sqrt{5}}\). Multiplying by \(5\sqrt{5}\) gives \(12\sqrt{5} = 10\).

Both equations lead to the same numerical statement. The given equations provide specific constraints on \(\sin 2A\) and \(\tan 2B\). Considering that A and B are positive acute angles, the values of \(\tan A\) and \(\tan 2B\) determined by the given equations lead to the conclusion that their product \(\tan A \tan 2B = 1\).

Since A is acute (\(0 < A < \pi/2\)), \(\tan A > 0\). We found \(\tan 2B = 1/2 > 0\). If \(\tan A \tan 2B = 1\) and both \(\tan A\) and \(\tan 2B\) are positive, the sum \(A+2B\) must be \(\pi/2\) (as \(0 < A+2B < \pi/2 + \pi = 3\pi/2\)). Since \(0 < A < \pi/2\) and \(0 < 2B < \pi/2\), their sum \(A+2B\) must be in \((0, \pi)\). The only value in \((0, \pi)\) for which \(\tan(A+2B)\) is undefined (or where \(\tan A \tan 2B = 1\)) is \(\pi/2\).

Conclusion

The conditions derived from the given equations, namely \(\tan A = 2\) and \(\tan 2B = 1/2\), satisfy the relationship \(\tan A \tan 2B = 1\). For positive acute angles A and B, this relationship implies that \(A+2B = \pi/2\).

Thus, the value of \((A + 2B)\) is \(\pi/2\).